Are homomorphisms into PGL related to the Schur multiplier?

I've been trying to understand homomorphisms from a finite group $G$ into $\operatorname{PGL}(n,R)$ for $n$ a positive integer, and $R$ a commutative ring with 1, usually a field.

I had been under the impression that these were induced by homomorphisms from the Darstellungsgruppen of $G$ into $\operatorname{GL}(n,R)$, but I am having doubts.

What is wrong with the following example?

Take $G$ to be cyclic of order 2. Its Schur multiplier is trivial, so it is its own Darstellungsgruppe.

For $r \in R^\times$, set $a=\begin{bmatrix} 0 & 1 \\ r & 0 \end{bmatrix} \in \operatorname{GL}(2,R)$. Then $a^2 = r a^0 \in Z(\operatorname{GL}(2,R))$ so $\bar a$ has order 2 in $\operatorname{PGL}(2,R)$.

Consider the homomorphism from $G$ to $\operatorname{PGL}(2,R)$ that sends a generator to $\bar a$. This is not induced by any homomorphism from a covering group of $G$ to $\operatorname{GL}(2,R)$ as long as $r$ has no square root in $R$.

Indeed any such homomorphism is determined by where it sends a generator, say to $b$, and to induce the original homomorphism to PGL $\bar b$ must equal $\bar a$.

If $\bar b = \bar a$ then $$b=\zeta a = \begin{bmatrix} 0 & \zeta \\ \zeta r & 0 \end{bmatrix} \quad b^2 = \begin{bmatrix} \zeta^2 r & 0 \\ 0 & \zeta^2 r \end{bmatrix}$$ If we want $b$ to have order 2, then $\zeta^2 r = 1$ and $\zeta^2 = \tfrac{1}{r}$, so if $r$ has no square root in $R^\times$, there is no such $b$.

Even if we claim the Darstellungsgruppen is cyclic of order 4 (since $H^2(C_2, R^\times)$ is likely equal to $(R^\times)/(R^\times)^2 \cong C_2$) we still need $\zeta^2r = -1$, but if $-1$ is a square in $R$, then $-r$ also has no square root, and it is still a counterexample.

We have $H^2(G,R^\times) \cong {\rm Hom}(M(G),R^\times) \oplus {\rm Ext}(G/G',R^\times)$, where $M(G)$ is the Schur multiplier of $G$.

If $R^\times$ is divisible (which is true, for example, if $R$ is algebraically closed), then ${\rm Ext}(G/G',R^\times) = 0$, and all extensions arises from a homomorphism from $M(G)$ to $R^\times$, and any homomorphism $G \to {\rm PGL}(2,R)$ lifts to $\hat{G} \to {\rm GL}(2,R)$ for some covering group $\hat{G}$ of $G$. But in general this will not be the case.

• If I really, really wanted a lift for my $G=C_2$, is this telling me I'd need to define $\hat G$ as an extension of the normal subgroup $R^\times$ with quotient $G/G'=C_2$? I think that roughly corresponds to taking a square root of $r$. Such a covering group is not very impressive, as it is not even finite. – Jack Schmidt Jun 18 '13 at 20:10
• (I think I roughly have it now, but I have to head out for a few hours. Will accept tomorrow after I have a chance to prove everything and write out the details.) – Jack Schmidt Jun 18 '13 at 20:28

The following is probably a misguided reinvention of Schur covers. It misses two key property of Schur covers: (1) Schur covers work for all projective representations (over an algebraically closed field simultaneously), and (2) Schur covers have a universal property that makes them minimal.

This answer also fails to prove Derek Holt's nice formula (which I assume is the universal coefficients theorem), which loses another important feature: if $G$ is a perfect group, the Schur cover is unique and projective representations (probably) lift over every commutative ring, not just algebraically closed fields.

At any rate, these results are enough to show that some sort of covering group exists, and if pressed further, some sort of covering group satisfying (1) is possible as well, though I have no idea if it will satisfy (2).

Statements of results

A representation group always exists for each projective representation. That is,


Proposition: We can always take $\ker(\phi) \cong \ker(\pi)$ if we'd like, but if $G$ is finite and $\ker(\pi)$ is $p$-divisible for each prime $p$ dividing the order of $G$, then we can take $\ker(\phi) \leq \{ k \in \ker(\pi) : k^{|G|} =1 \}$.

Corollary: In particular, if $H=\PGL(n,R)$ and $\hat H=\GL(n,R)$ for $R$ an integrally closed order in an algebraically closed field, then we can take $\ker(\phi)$ finite and cyclic.

Proofs

Proof (of first proposition): We begin with the image of $\hat \rho$: For each $g \in G$, choose some preimage $\mu(g) \in \hat H$ of $\rho(g)$ under $\pi$. Thus $\mu:G \to \hat H$ is a function satisfying $\pi(\mu(g)) = \rho(g)$. We define $$\zeta(g,h) = \mu(gh)^{-1} \mu(g) \mu(h)$$ Consider $\pi(\zeta(g,h)) = \pi(\mu(gh))^{-1} \pi(\mu(g)) \pi(\mu(h)) = \rho(gh)^{-1} \rho(g) \rho(h) = \rho(1) =1$, so $$\zeta : G^2 \to \ker(\pi)$$ Now consider $$\zeta(gh,k)\zeta(g,h) = \mu(ghk)^{-1} \mu(gh) \mu(k) \mu(gh)^{-1} \mu(g) \mu(h) = \mu(ghk)^{-1} \mu(g)\mu(h)\mu(k)$$ versus $$\zeta(g,hk)\zeta(h,k) = \mu(ghk)^{-1} \mu(g) \mu(hk) \mu(hk)^{-1} \mu(h) \mu(k) = \mu(ghk)^{-1} \mu(g) \mu(h) \mu(k)$$ hence we get $$\zeta(gh,k) \zeta(g,h) = \zeta(g,hk)\zeta(h,k) \qquad \zeta \in Z^2(G,\ker(\pi))$$ This allows us to define a group $\hat G$ on the set $G \times K$ for any subgroup $K \leq \ker(\pi)$ that contains the image of $\zeta$ by $$\hat G: \qquad (g,k)\cdot (h,l) = (gh,kl\cdot \zeta(g,h))$$ The cocycle condition, $Z^2(G,\ker(\pi))$ is precisely what is needed for $\hat G$ to be a group, that is, for the multiplication to be associative. Define $$\hat \rho :\hat G \to \hat H: (g,k) \mapsto \mu(g) k \qquad \phi:\hat G \to G:(g,k) \mapsto g$$ We calculate that $$\hat \rho( (g,k) \cdot (h,l) ) = \hat \rho((gh,kl\zeta(g,h)) = \mu(gh) kl \mu(gh)^{-1} \mu(g)\mu(h) = \mu(g)k \cdot \mu(h)l = \hat \rho(g,k) \hat \rho(h,l)$$ Hence $\hat \rho$ is a homomorphism. Now we check that $$\pi(\hat \rho((g,k))) = \pi(\mu(g)k) = \pi(\mu(g)) \pi(k) = \rho(g) \cdot 1 = \rho(g) =\rho(\phi((g,k))$$ Hence the diagram commutes as was to be shown. $\square$

Lemma: Define $\tau(h) = \prod_{g \in G} \zeta(g,h)$ for $\zeta \in Z^2(G,K)$. Then $\tau(gh)^{-1} \tau(g) \tau(h) = {\zeta(g,h)}^{|G|}$.

Proof (of lemma): Notice $$\begin{array}{rl} {\left(\zeta(h,k)\right)}^{|G|} \cdot \tau(hk) &= \prod_{g \in G} \left( \zeta(g,hk) \zeta(h,k) \right) \\ &= \prod_{g \in G} \left( \zeta(gh,k) \zeta(g,h) \right) \\ &= \left( \prod_{g \in G} \zeta(gh,k) \right) \cdot \left( \prod_{g \in G} \zeta(g,h) \right) \\ & = \tau(k) \cdot \tau(h) \end{array}$$ where the equality for $\tau(k)$ uses the fact that $\{ gh : g \in G \} = \{ g : g \in G \}$. $\square$

Proof (of second proposition): Set $\tau(g)$ as in the lemma, and for each $g$, let $\bar \tau(g)$ be an $n$th root of $\tau(g)$. Set $\bar \mu(g) = \mu(g) / \bar \tau(g)$ and so $$\bar \zeta(g,h) = \mu(gh)^{-1} \mu(g) \mu(h) \bar \tau(gh) \bar \tau(g)^{-1} \bar \tau(h)^{-1}$$ More importantly $$\begin{array}{rl} {\left( \bar \zeta(g,h) \right)}^{|G|} &= {\left(\zeta(g,h)\right)}^{|G|} \cdot {\left( \bar \tau(gh)\right)}^{|G|} \cdot {\left( \bar \tau(g) \right)}^{-|G|} \cdot {\left( \bar \tau(h) \right)}^{-|G|} \\ &= {\left(\zeta(g,h)\right)}^{|G|} \cdot \tau(gh) \tau(g)^{-1} \tau(h)^{-1} \\ &= {\left(\zeta(g,h)\right)}^{|G|} \cdot {\left(\zeta(g,h)\right)}^{-|G|} \\ &= 1 \end{array}$$ It is standard and not hard to check that $\zeta$ and $\bar \zeta$ define isomorphic $\hat G$ if we use the same $K$. The advantage now is that we can view $\bar \zeta \in Z^2(G,K)$ for $K \leq \{ k \in \ker(pi) : k^n = 1\}$. By definition of $\hat G$ and $\phi$, one always has $\ker(\phi)\cong K$, so we are done. $\square$

Proof (of the corollary): The kernel of $\pi$ is exactly $R^\times$. Since $R \leq K$ and $K$ is algebraically closed, $K^\times$ is divisible, so an $n$th root of $\tau(h)$ always exists. Since $R$ is integrally closed in $K$, that $n$th root lies in $R$. Since $\tau(h) \in R^\times$ is a unit, its divisor $\bar \tau(h) \in R^\times$ must also be a unit. Hence the second proposition applies to choose $K \leq \{ r \in R^\times : r^{|G|} =1 \}$, the group of $n$th roots of unity, a finite cyclic group. $\square$